Question 163488
number of zeros = number of factors of 5 in the product


50 * 51 * ... * 200 = (200!)/(49!)


number of zeros in 200! = 200/5 + 200/5^2 + ... = 49 (must be an integer)

number of zeros in 49! = 49/5 + 49/5^2 + ... = 10 (must be an integer)


number of zeros in 50 * 51 * ... * 200 = 49 - 10 = 39 zeroes